Anyone?

$\large{\displaystyle \int^{c}_{b} \frac{\ln x}{\sqrt{x}(x+a)}dx}$

#Calculus

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Comments

Well I am getting this -

$\displaystyle \dfrac{1}{a} \left[\frac{\sqrt{z} log(z)}{z + a} + \frac{\sqrt{z} log(z)}{2a} \Phi\left(-\frac{z}{a}, 1, \frac{1}{2}\right) - \frac{\sqrt{z}}{a} \Phi\left(-\frac{z}{a}, 1, \frac{1}{2}\right)\right]_b^c$

Kartik Sharma - 5 years, 9 months ago

ok frankly i didnt understand what u have typed here. anyway i was trying to find

$\large{\displaystyle \int^{16}_{1} \frac{\ln x}{\sqrt{x}(x+4)}dx}$

Tanishq Varshney - 5 years, 9 months ago

Substitute $x = y^2$ . Then put $y=\dfrac{2}{t}$ to again get the same integral and evaluate it. I'll post the complete solution if you want.

Sudeep Salgia - 5 years, 9 months ago

@Sudeep Salgia – yup plz

Tanishq Varshney - 5 years, 9 months ago

@Tanishq Varshney – Let $\displaystyle I = \int_1^{16} \frac{ \ln x }{\sqrt{x}(x+4) } dx$ . Substitute $x = y^2$ to get $\displaystyle I = \int_1^{4} \frac{ 4 \ln y }{(y^2+4) } dy$ . Now put $y = \dfrac{4}{t}$ to get $\displaystyle \frac{I}{4} = \int_1^4 \frac{ \ln 4 - \ln t}{(4 + t^2)} dt$ . Rearranging, we get, $\displaystyle I = 2 \ln 4 \int_1^4 \frac{1}{t^2 + 4} dt$ .
This is a simple integration. So the answer becomes, $I = \ln 4 ( \arctan (2) - \arctan(1/2))$

Hope this helps. Please check and let me know if there are any calculation mistakes. (I wasn't writing any thing down :P )

Sudeep Salgia - 5 years, 9 months ago

@Sudeep Salgia – Clever substitution!

Aditya Kumar - 5 years, 9 months ago

@Sudeep Salgia – Can you please explain the method where you divide I/4 . And when you re-arrange ln (t) disappears ?

Thanks.

Syed Baqir - 5 years, 9 months ago

But I need to check once again.

Kartik Sharma - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$